33

我有两个列表，其元素的名称部分重叠，我需要将它们逐个元素合并/组合成一个列表：

> lst1 <- list(integers=c(1:7), letters=letters[1:5],
                words=c("two", "strings"))
> lst2 <- list(letters=letters[1:10], booleans=c(TRUE, TRUE, FALSE, TRUE),
                words=c("another", "two"), floats=c(1.2, 2.4, 3.8, 5.6))

> lst1
$integers
[1] 1 2 3 4 5 6 7

$letters
[1] "a" "b" "c" "d" "e"

$words
[1] "two"     "strings"

> lst2
$letters
 [1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j"

$booleans
[1]  TRUE  TRUE FALSE  TRUE

$words
[1] "another" "two"    

$floats
[1] 1.2 2.4 3.8 5.6

我尝试使用mapply，它基本上按索引组合了两个列表（即：“[[”），而我需要按名称组合它们（即：“$”）。此外，由于列表具有不同的长度，因此应用了回收规则（结果相当不可预测）。

> mapply(c, lst1, lst2)
$integers
 [1] "1" "2" "3" "4" "5" "6" "7" "a" "b" "c" "d" "e" "f" "g" "h" "i" "j"

$letters
[1] "a"     "b"     "c"     "d"     "e"     "TRUE"  "TRUE"  "FALSE" "TRUE" 

$words
[1] "two"     "strings" "another" "two"    

$<NA>
 [1] 1.0 2.0 3.0 4.0 5.0 6.0 7.0 1.2 2.4 3.8 5.6

Warning message:
In mapply(c, lst1, lst2) :
  longer argument not a multiple of length of shorter

正如您可能想象的那样，我正在寻找的是：

$integers
[1] 1 2 3 4 5 6 7

$letters
[1] "a" "b" "c" "d" "e" "a" "b" "c" "d" "e" "f" "g" "h" "i" "j"

$words
[1] "two"     "strings"   "another" "two"

$booleans
[1]  TRUE  TRUE FALSE  TRUE

$floats
[1] 1.2 2.4 3.8 5.6

有没有办法做到这一点？谢谢！

4

4 回答 4

45

你可以做：

keys <- unique(c(names(lst1), names(lst2)))
setNames(mapply(c, lst1[keys], lst2[keys]), keys)

泛化到任意数量的列表需要混合do.call和lapply：

l <- list(lst1, lst2, lst1)
keys <- unique(unlist(lapply(l, names)))
setNames(do.call(mapply, c(FUN=c, lapply(l, `[`, keys))), keys)
于 2013-08-30T18:20:47.650 回答
1

弗洛德尔对用户的回答更新tidyverse：

list1 <- list(integers=c(1:7), letters=letters[1:5],
               words=c("two", "strings"))
list2 <- list(letters=letters[1:10], booleans=c(TRUE, TRUE, FALSE, TRUE),
               words=c("another", "two"), floats=c(1.2, 2.4, 3.8, 5.6))

input_list <- list(list1, list2, list1, list2)

我们希望为输出列表中的每个元素精确地复制原始所需的输出两次。使用map2and ，我们可以比涉及,和reduce的基本解决方案更清楚地实现这一点。首先，我们声明了一个函数，该函数通过使用它们的命名元素组合两个列表，然后我们通过以下方式在输入列表上调用我们的函数：Rdo.callmapplylapplyc()reduce

library(purrr)

cat_lists <- function(list1, list2) {  

  keys <- unique(c(names(list1), names(list2)))
  map2(list1[keys], list2[keys], c) %>% 
    set_names(keys)  

}

combined_output <- reduce(input_list, cat_lists)

这给了我们想要的东西：

> combined_output

#> $integers
#>  [1] 1 2 3 4 5 6 7 1 2 3 4 5 6 7
#> 
#> $letters
#>  [1] "a" "b" "c" "d" "e" "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "a" "b"
#> [18] "c" "d" "e" "a" "b" "c" "d" "e" "f" "g" "h" "i" "j"
#> 
#> $words
#> [1] "two"     "strings" "another" "two"     "two"     "strings" "another"
#> [8] "two"    
#> 
#> $booleans
#> [1]  TRUE  TRUE FALSE  TRUE  TRUE  TRUE FALSE  TRUE
#> 
#> $floats
#> [1] 1.2 2.4 3.8 5.6 1.2 2.4 3.8 5.6
于 2019-07-23T20:34:53.300 回答
0

我也用grep，不知道是更好，最差，还是等价的！

l_tmp <- c(lst1, lst2, lst1)
keys = unique(names(l_tmp))
l = sapply(keys, function(name) {unlist(l_tmp[grep(name, names(l_tmp))])})
于 2015-08-25T11:57:10.407 回答
0

我将添加我自己的基于该tapply功能的解决方案。

lst1 <- list(integers=c(1:7), letters=letters[1:5],
               words=c("two", "strings"))
lst2 <- list(letters=letters[1:10], booleans=c(TRUE, TRUE, FALSE, TRUE),
               words=c("another", "two"), floats=c(1.2, 2.4, 3.8, 5.6))

binded <- c(lst1, lst2) # and for list of lists Reduce("c", list(lst1, lst2))

tapply(binded, names(binded), function(x) unlist(x, FALSE, FALSE)) # double false for better performance
于 2021-12-12T08:56:51.003 回答